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Re: Cann't print local vars when nesting functions


Eli Zaretskii wrote:

Date: Fri, 18 Feb 2005 17:54:17 -0500
From: Jeff <wd4nmq@comcast.net>

int main(int argc, char *argv[]){
   int i,j;

   int inside(void){
      int k,l;
      k = 3;
      l = 4;
   }

   i = 0;
   j = 1;
}

Now set the break point to the same line, j =1, and this happens:
Breakpoint 1, main () at test.c:25
25        j = 1;
(gdb) p i
No symbol "i" in current context.
(gdb)

What gives? Is there something you need to special when nesting functions?



Please tell the details: what platform is this, what versions of GCC and GDB you use, and how (with what command-line options) you compiled and linked the program. Also, since the line "j = 1;" is not line 25 in the source you posted, could it be that the program you actually compiled was different?

FWIW, I tried this with GCC 3.3.3, naive compiler command line, and
GDB 6.3, and couldn't reproduce the problem with the source you
posted.



I am on linux, 2.4.26 kernel
gcc 3.2.3
gdb 5.3
Source file test.c:
#include <stdio.h>

int main(int argc, char *argv[]){

 int i,j;
 int inside(void){
   int k,l;

   k = 1;
   l = k;
   return 0;

}

 i = 0;
 j = 1;
}

test$ cc -g -O0 test.c
$ gdb ./a.out
GNU gdb 5.3
Copyright 2002 Free Software Foundation, Inc.
GDB is free software, covered by the GNU General Public License, and you are
welcome to change it and/or distribute copies of it under certain conditions.
Type "show copying" to see the conditions.
There is absolutely no warranty for GDB. Type "show warranty" for details.
This GDB was configured as "i386-slackware-linux"...
(gdb) b test.c:16
Breakpoint 1 at 0x8048340: file test.c, line 16.
(gdb) run
Starting program: test/a.out


Breakpoint 1, main () at test.c:16
16        j = 1;
(gdb) p i
No symbol "i" in current context.
(gdb)

As stated, removing the inside() function, breaking on the same source line and printing causes i's value to be printed. No error message.

tj



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